Vector

Let $\QTR{bf}{R}^3$ denote the set of all triples with Elements in $\QTR{bf}{R}^3$ are called vectors. The sum of vectors

is

The scalar multiplication of with the vector is

The zero vector is denoted For is defined by These rules have properties

for
for
for
for
for
for
for $\alpha ,\beta$ $\in \QTR{bf}{R,}$ for
for $\alpha ,\beta$ $\in \QTR{bf}{R,}$ for

Geometric Interpretation

translation: $x\rightarrow x+a$

magnification: $\alpha >1$

contraction:

parallelogram construction

Dot Product

The dot product of two vectors and is

Proposition

for
$a\cdot b=b\cdot a$ for
for
for
for
for
$a\cdot a>0$ unless for (Proof. )

Cauchy-Schwarz Inequality

The equality holds if and only if there exist with

Proof. Case 1. Then so the equality holds.

Case 2. $a\ne 0.$ Then $a\cdot a>0.$ Consider the vector

Then

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Therefore

If and $a\ne 0,$ then so i.e. If and $b\ne 0,$ then

The norm of the vector is
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It follows from Cauchy-Schwarz inequality that
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if $a\ne 0,b\ne 0.$ The unique angle such that
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is called the angle between the vectors and This is consistent with the Law of Cosine:

Properties of the Norm

$\Vert a\Vert \ge 0$ for all $\Vert a\Vert =0$ if and only if
for
Triangle Inequality: for

Proof.

for
Parallelogram Law: for

Proof. Add the identities

Two vectors are perpendicular $(a\perp b)$ if the angle between them equals $\pi /2:$
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Projections and Components

Theorem. Let $a\ne 0.$ Each $b\in \QTR{bf}{R}^3$ can be uniquely expressed as

where $b_{\Vert }$ is parallel to (i.e., there exists with and $b_{\perp }$ is perpendicular to (i.e.,

Proof. Set and Then

If with then

so ; this shows and

Definition is called the projection of the vector on the vector

Direction Cosine

Notation:

If $a\ne 0,$ then the vector satisfies i.e., is a unit vector. The components of in the directions

are called the direction cosines of

Cross Product

Let Their cross product (also called vector product) is given by
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Properties of Cross Product

$a\times a=0$ for
for
for
for
for $\alpha ,\beta$ $\in \QTR{bf}{R,}$ for
Scalar Triple Product:

Proof.

Therefore
for
Hence $a\times b$ is perpendicular to both and
Lagrange Identity: for

Proof.
is the area of the parallelogram formed with sides and

Proof. We may assume that $a\ne 0$ and $b\ne 0.$ From the identity

we see that

where $\theta$ is the angle between and It follows that

therefore the required area equals
The absolute value of the triple product represents the volume of the parallelepiped with edges

Proof. Write

The quantity

is the component of the vector along the direction so the height of the parallelepiped with base formed by the vectors is The result follows from the fact
CAB Minus BAC Identity

The result holds for $a,b,c\in \{i,j,k\}$ (there are altogether 27 different cases.) Write

The general case follows from
Extended Lagrange Identity:

Proof.

It follows from CAB minus BAC identity that this equals

Note that the Lagrange identity may be obtained from the extended Lagrange identity by setting and
Jacobi Identity:

Proof. Add the identities

Cramer's Rule

Consider the simultaneous set of equations

From the basic properties of the determinant we have
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Hence if is to satisfy equations (1) then
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Write
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If $\Delta \ne 0,$ then
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Similarly, if $\Delta \ne 0,$ then
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On the other hand, the following identity always holds:
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This four-by-four determinant can be expanded along the first row:
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Hence if $\Delta \ne 0,$ then
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is indeed the (unique) solution to the system (1).

The method can be extended to linear equations in unknowns.

Straight Line in $\QTR{bf}{R^3}$

A straight line in $\QTR{bf}{R^3}$ can be given by the parametric equations:
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A point belong to this line if and only if

for some $t\in \QTR{bf}{R.}$ This can therefore be written in vector equations as:
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After eliminating the parameter from the parametric equation, we obtain the symmetric equation
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if $abc\ne 0.$

Example Find the equations of the straight line passing through and .

First we determine the direction: from this the vector equation is obtained:

Equating the components we obtain the parametric equation:
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and the symmetric equations:
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Example Find the distance between the skew lines given by
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and given by
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We note that has direction and has direction The vector
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is perpendicular to both and Choose any point $p\in L_1$ and any point $q\in L_2.$ (There are infinitely many ways to do so.) The distance between and is the norm of the projection of on Take and Then and so the required distance is
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Plane in $\QTR{bf}{R^3}$

Let be any point of the plane. The plane consists of all the points satisfying the vector equation

for a nonzero vector This vector is called the normal of the plane. Thus the equation of the plane takes the form

where

Example Find the distance from to the plane

The vector is normal to the plane. Choose a point on the plane: the distance from to the plane is the norm of the projection of the vector on
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Example Show that the line
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can be expressed as the intersection of two planes, each of which is parallel to a coordinate axis.

Consider the plane with equation obtained from the first equality
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can be expressed by the equation

which has as normal. is therefore parallel to the -axis since is perpendicular to Similarly, the plane satisfying the equation
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which is the same as the equation

has as normal, must be parallel to the -axis since is perpendicular to The given line is the intersection of and

Intersecting Planes

The angle between two intersecting planes is the same as the angle between their normals . Depending on the choices of the normals, there are two such angles, each the supplement of the other. We choose the smaller angle, the one with the nonnegative cosine:
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Example Find the cosine of the angle between the planes with equations

and

Solution: ; so
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Example The planes

and

intersect to form a line. Find the vector equation of this line.

Solution: The direction vector of the intersecting line must be perpendicular to both normal vectors and Hence it is parallel to Let be any point on the intersecting line. The required vector equation takes the form

Plane Determined by Three Noncollinear Points

Suppose that the points

do not lie on a straight line. A point lies on the plane determined by these three points if and only if the volume of the parallelepiped with as vertices is zero. In view of the vector triple product identity, the condition is given by

Example Find the equation for the plane passing through

Solution: Expanding the determinant
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we have

Distance from a Point to a Plane

Suppose that does not lie on the plane

Choose any point on the plane. The distance from to the plane is the norm of the projection of on the normal of the plane:
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Since lies on the plane, it follows that

Therefore

consequently the required distance is given by
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Exercise

Find the equation for the plane
1. through perpendicular to
2. through perpendicular to
3. through parallel to
4. through parallel to
Explain why a plane cannot
1. contain and and be perpendicular to
2. contain and
3. go through the origin and have the equation
Find the projection of along and also find $\Vert P\Vert$
3. unit vector at $60^{\circ }$ angle with
4. vector of length at $60^{\circ }$ angle with
7. is perpendicular to
8. is perpendicular to
True or false:
1. $A\times B$ never equals $A\cdot B.$
2. If $A\times B=0$ and $A\cdot B=0,$ then either or
3. If and $A\ne 0,$ then
Which of the following equals to $A\times B?$
1. $(A+B)\times B$
2. $(-B)\times (-A)$
4. $(A+C)\times (B-C)$